Question 15.3: First-Order Rate Laws II Using the data given in Example 15....
First-Order Rate Laws II
Using the data given in Example 15.2, calculate [N_{2}O _{5}] 150. s after the start of the reaction.
We know from Example 15.2 that [N_{2}O _{5}] = 0.0500 mol/L at 100 s and [N_{2}O _{5}] = 0.0250 mol/L at 200 s. Since 150 s is halfway between 100 s and 200 s, it is tempting to assume that we can use a simple average to obtain [N_{2}O _{5}] at that time. This is incorrect because it is \ln [N_{2}O _{5}] , not [N_{2}O _{5}] , that depends directly on t. To calculate [N_{2}O _{5}] after 150 s, we must use Equation (15.2):
\ln [A] = -k t + \ln [A]_{0} (15.2)
\ln [N_{2}O _{5}] = -k t + \ln [N_{2}O _{5}]_{0}
where t = 150. s, k = 6.93 × 10 ^{-3} s ^{-1} (as determined in Example 15.2), and [N_{2}O _{5}]_{0} = 0.100 mol/L.
\ln( [N_{2}O _{5}])_{t= 150} = – (6.93 × 10 ^{-3} s ^{-1}) (150. s) + \ln( 0.100)
= – 1.040 – 2.303 = – 3.343
( [N_{2}O _{5}])_{t= 150} = antilog ( – 3.343) = 0.0353 mol / L
Note that this value of [N_{2}O _{5}] is not halfway between 0.0500 mol/L and 0.0250 mol/L.